Question

Let $PQR$ be a triangle of area $\Delta$ with $a=2,b=\frac{7}{2}$ and $c=\frac{5}{2}$, where $a,b$ and $c$ are the lengths of the sides of the triangle opposite to the angles at $P,Q$ and $R$ respectively. Then $\frac{2\sin P-\sin 2P}{2\sin P+\sin 2P}$ equals

• A. $\frac{3}{4\Delta }$
• B. $\frac{45}{4\Delta }$
• C. ${\left(\frac{3}{4\Delta }\right)}^2$
• D. ${\left(\frac{45}{4\Delta }\right)}^2$

Answer 1

The correct option is C ${\left(\frac{3}{4\Delta }\right)}^2$

$\frac{2sinP-2sinPcosP}{2sinP+2sinPcosP}=\frac{1-cosP}{1+cosP}$

$=\frac{2si{n}^2\frac{P}{2}}{2co{s}^2\frac{P}{2}}=ta{n}^2\frac{P}{2}$

$=\frac{(s-b)(s-c)}{s(s-a)}$

$=\frac{\left(\right(s-b\left)\right(s-c){)}^2}{{\Delta }^2}[\because \Delta =\sqrt{s(s-a)(s-b)(s-c)}]$

$=\frac{\left(\right(\frac{1}{2}\left)\right(\frac{3}{2}){)}^2}{\Delta }[s=\frac{a+b+c}{2}]$

$=(\frac{3}{4\Delta }{)}^2$